Finding Closest Point on Plane to (5,6,7)

[cleopatra]

Homework Statement

I have a plane: 20x+15y+10z=4
Find the point on the plane where it´s closest to the point (5,6,7)

Homework Equations

any ideas?

The Attempt at a Solution

No idea!

 

[jambaugh]

Is this a calculus problem or a linear algebra problem?

If a calculus problem express the distance between the point and an arbitrary point on the plane as a function of x,y,z. Then impose the conditions on the plane to reduce to two variables. Then you optimize the distance (or more easily the square of the distance) which will occur when the partial derivatives are zero.

If a linear algebra or analytic geometry problem then note that the point on the plane where this is the case will be such that the line through it and the point (5,6,7) will be orthogonal to the plane and thus parallel to the vector <20,15,10>.

 

[cleopatra]

It´s a linear algebra problem and I don´t know how to do it. I have no formulas at all.

 

[jambaugh]

The coefficients of the linear equation defining the plane are coordinates of a vector normal (perpendicular) to the plane.

Call this vector N.

Now express a vector pointing from the given point to an arbitrary point (x,y,z) on the plane. That'd be V= <5-x,6-y,7-z>.

Now solve the system of equations V = cN along with the equation for the plane. The solution will give x,y, and z for the point you're looking for and the magnitude of V will give you the distance.

 

[cleopatra]

cN=V
then
c(20,15,10)=(5-x, 6-y, 7-z)

Then what?

20c=5-x
15c=6-y
10c=7-z
?

 

[jambaugh]

cN=V
then
c(20,15,10)=(5-x, 6-y, 7-z)

Then what?

20c=5-x
15c=6-y
10c=7-z
?

Right, but also include the equation requiring (x,y,z) lie on the plane:
20x + 15y + 10z = 4.
That's 4 equations and 4 unknowns c,x,y, and z.
I suggest you solve the three for x,y, and z in terms of c then substitute those into the plane equation to find c.

 

[HallsofIvy]

What are parametric equations of the line through (5, 6, 7) with direction vector <10, 15, 10>? Where does that line intersect 20x+ 15y+ 10z= 4?

(Since the length of the direction vector is irrelevant to the line, the "c" in cN= V can be anything. I am suggesting you take it to be equal to 1.)

 

[cleopatra]

Right, but also include the equation requiring (x,y,z) lie on the plane:
20x + 15y + 10z = 4.
That's 4 equations and 4 unknowns c,x,y, and z.
I suggest you solve the three for x,y, and z in terms of c then substitute those into the plane equation to find c.

I´m sorry, I don´t understand you :/

 

[jambaugh]

I´m sorry, I don´t understand you :/

You've got three equations which you can rewrite:
x = 5 - 20c
y = 6 - 15c
z = 7 - 10c

you've got the equation for the plane:
20x + 15y + 10z = 4.

Substitute (replace x in the plane equation with (5-20c) and so on) and you have an equation you can solve for the value c.
when you know c you know x, y, and z, which are the coordinates of the point you are looking for.

 

[sciencegirl1]

What I need is the way to solve this problem. Step by step.
I don´t need to understand it. I can´t see why you don´t want to help me.

 

[sutupidmath]

What I need is the way to solve this problem. Step by step.
I don´t need to understand it. I can´t see why you don´t want to help me.

It is a forum rule, not to simply give out answers. So, you better try to understand it for people here won't do the homework for you!